Question: $\text E = \left[\begin{array}{rr}3 & 5 \\ -1 & 1\end{array}\right]$ and $\text A = \left[\begin{array}{rrr}-2 & 2 & 3 \\ 3 & 5 & -2\end{array}\right]$ Let $\text {H = EA}$. Find $\text H$. $ {H = }$
The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{E}$ and the first column of $\text{A}$. $ \text {H}=\left[\begin{array}{rr}{3} & {5} \\ -1 & 1\end{array}\right]\left[\begin{array}{rr} {-2} & 2 & 3 \\ {3} & 5 & -2\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(3,5)\cdot(-2,3)\\\\ &=3 \cdot -2 + 5\cdot 3\\\\ &=9 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1 \cdot 2 + 1\cdot 5 = 3$ (Choice B) B $-1 \cdot -2 + 1\cdot 3 = 5$ (Choice C) C $3 \cdot 2 + 5\cdot 5 = 31$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rrr}9 & 31 & -1 \\ 5 & 3 & -5\end{array}\right]$